An element of occurs in bcc structure with the cell edge 288 pm its density is 7.8 g/cm^3 . calculate the atomic mass of the element
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Volume of unit cell =(288 pm)3=(288×10−10 cm)3=2.389×10−23 cm3
Volume of 208 g of the element =DensityMass=7.2208=28.89 cm3
Number of unit cells =Volume of a unit cellTotal Volume=2.389×10−2328.89=12.09×1023
For a BCC structure, number of atoms per unit cell =2
∴ Number of atoms present in 208 g = No. of atoms per unit cell × No. of unit cells
=2×12.09×1023
=24.18×1023
=2.418×1024
Explanation:
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